Define $a_n$ as follows:
$$ a_1=1,\ \ a_{n+1}=na_n+1\ $$ At this time, the sequence $a_n$ is as follows: $$ a_n=\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!} $$ I made some discoveries about this sequence.
The first:$$a_k\equiv 0\pmod{m}\Rightarrow a_{k+Nm}\equiv 0\pmod{m}~~~~\forall k,m,N\in\mathbb{N}$$ The second:$$ n\geq 4\,\Rightarrow\,a_n ~\mathrm{is~composite} $$ I was able to prove the first, but not the second. My expectation is that the second is correct, but I'm not sure it can be proved. Please let me know if you come up with a proof method. Any help is welcome!

(I am a Japanese college student. I'm sorry for my poor English.)

Define $a_n$ as follows:
$$ a_1=1,\ \ a_{n+1}=na_n+1\ $$ At this time, the sequence $a_n$ is as follows: $$ a_n=\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!} $$ I made some discoveries about this sequence.
The first:$$a_k\equiv 0\pmod{m}\Rightarrow a_{k+Nm}\equiv 0\pmod{m}~~~~\forall k,m,N\in\mathbb{N}$$ The second:$$ n\geq 4\,\Rightarrow\,a_n ~\mathrm{is~composite} $$ I was able to prove the first, but not the second. My expectation is that the second is correct, but I'm not sure it can be proved. My friend used computer and check $a_n$ is composite for $4\leq n\leq 48$. After $a_{49}$, it is too large number to check on his computer. Please let me know if you come up with a proof method. Any help is welcome!

(I am a Japanese college student. I'm sorry for my poor English.)